Assuming the order required is as n-> inf.
As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.
We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).
So yes, both are o(log(n)).
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